Problem Transformed CSR 04 Ex5 Zan97 C

Tool CaT

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 C

stdout:

YES(?,O(n^1))

Problem:
active(f(X)) -> mark(if(X,c(),f(true())))
active(if(true(),X,Y)) -> mark(X)
active(if(false(),X,Y)) -> mark(Y)
active(f(X)) -> f(active(X))
active(if(X1,X2,X3)) -> if(active(X1),X2,X3)
active(if(X1,X2,X3)) -> if(X1,active(X2),X3)
f(mark(X)) -> mark(f(X))
if(mark(X1),X2,X3) -> mark(if(X1,X2,X3))
if(X1,mark(X2),X3) -> mark(if(X1,X2,X3))
proper(f(X)) -> f(proper(X))
proper(if(X1,X2,X3)) -> if(proper(X1),proper(X2),proper(X3))
proper(c()) -> ok(c())
proper(true()) -> ok(true())
proper(false()) -> ok(false())
f(ok(X)) -> ok(f(X))
if(ok(X1),ok(X2),ok(X3)) -> ok(if(X1,X2,X3))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Proof:
Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {10,9,8,7,6}
   transitions:
    top1(18) -> 10*
    active1(5) -> 18*
    active1(2) -> 18*
    active1(4) -> 18*
    active1(1) -> 18*
    active1(3) -> 18*
    proper1(5) -> 18*
    proper1(2) -> 18*
    proper1(4) -> 18*
    proper1(1) -> 18*
    proper1(3) -> 18*
    ok1(15) -> 18,9
    ok1(14) -> 14,8
    ok1(11) -> 11,7
    if1(5,2,1) -> 14*
    if1(3,2,2) -> 14*
    if1(5,4,1) -> 14*
    if1(3,4,2) -> 14*
    if1(2,1,5) -> 14*
    if1(2,3,5) -> 14*
    if1(2,5,5) -> 14*
    if1(1,1,4) -> 14*
    if1(1,3,4) -> 14*
    if1(1,5,4) -> 14*
    if1(5,1,3) -> 14*
    if1(5,3,3) -> 14*
    if1(5,5,3) -> 14*
    if1(4,1,2) -> 14*
    if1(1,1,1) -> 14*
    if1(4,3,2) -> 14*
    if1(1,3,1) -> 14*
    if1(4,5,2) -> 14*
    if1(1,5,1) -> 14*
    if1(3,2,5) -> 14*
    if1(3,4,5) -> 14*
    if1(2,2,4) -> 14*
    if1(2,4,4) -> 14*
    if1(1,2,3) -> 14*
    if1(1,4,3) -> 14*
    if1(5,2,2) -> 14*
    if1(2,2,1) -> 14*
    if1(5,4,2) -> 14*
    if1(2,4,1) -> 14*
    if1(4,1,5) -> 14*
    if1(4,3,5) -> 14*
    if1(4,5,5) -> 14*
    if1(3,1,4) -> 14*
    if1(3,3,4) -> 14*
    if1(3,5,4) -> 14*
    if1(2,1,3) -> 14*
    if1(2,3,3) -> 14*
    if1(2,5,3) -> 14*
    if1(3,1,1) -> 14*
    if1(1,1,2) -> 14*
    if1(3,3,1) -> 14*
    if1(1,3,2) -> 14*
    if1(3,5,1) -> 14*
    if1(1,5,2) -> 14*
    if1(5,2,5) -> 14*
    if1(5,4,5) -> 14*
    if1(4,2,4) -> 14*
    if1(4,4,4) -> 14*
    if1(3,2,3) -> 14*
    if1(3,4,3) -> 14*
    if1(4,2,1) -> 14*
    if1(2,2,2) -> 14*
    if1(4,4,1) -> 14*
    if1(2,4,2) -> 14*
    if1(1,1,5) -> 14*
    if1(1,3,5) -> 14*
    if1(1,5,5) -> 14*
    if1(5,1,4) -> 14*
    if1(5,3,4) -> 14*
    if1(5,5,4) -> 14*
    if1(4,1,3) -> 14*
    if1(4,3,3) -> 14*
    if1(4,5,3) -> 14*
    if1(5,1,1) -> 14*
    if1(3,1,2) -> 14*
    if1(5,3,1) -> 14*
    if1(3,3,2) -> 14*
    if1(5,5,1) -> 14*
    if1(3,5,2) -> 14*
    if1(2,2,5) -> 14*
    if1(2,4,5) -> 14*
    if1(1,2,4) -> 14*
    if1(1,4,4) -> 14*
    if1(5,2,3) -> 14*
    if1(5,4,3) -> 14*
    if1(4,2,2) -> 14*
    if1(1,2,1) -> 14*
    if1(4,4,2) -> 14*
    if1(1,4,1) -> 14*
    if1(3,1,5) -> 14*
    if1(3,3,5) -> 14*
    if1(3,5,5) -> 14*
    if1(2,1,4) -> 14*
    if1(2,3,4) -> 14*
    if1(2,5,4) -> 14*
    if1(1,1,3) -> 14*
    if1(1,3,3) -> 14*
    if1(1,5,3) -> 14*
    if1(5,1,2) -> 14*
    if1(2,1,1) -> 14*
    if1(5,3,2) -> 14*
    if1(2,3,1) -> 14*
    if1(5,5,2) -> 14*
    if1(2,5,1) -> 14*
    if1(4,2,5) -> 14*
    if1(4,4,5) -> 14*
    if1(3,2,4) -> 14*
    if1(3,4,4) -> 14*
    if1(2,2,3) -> 14*
    if1(2,4,3) -> 14*
    if1(3,2,1) -> 14*
    if1(1,2,2) -> 14*
    if1(3,4,1) -> 14*
    if1(1,4,2) -> 14*
    if1(5,1,5) -> 14*
    if1(5,3,5) -> 14*
    if1(5,5,5) -> 14*
    if1(4,1,4) -> 14*
    if1(4,3,4) -> 14*
    if1(4,5,4) -> 14*
    if1(3,1,3) -> 14*
    if1(3,3,3) -> 14*
    if1(3,5,3) -> 14*
    if1(4,1,1) -> 14*
    if1(2,1,2) -> 14*
    if1(4,3,1) -> 14*
    if1(2,3,2) -> 14*
    if1(4,5,1) -> 14*
    if1(2,5,2) -> 14*
    if1(1,2,5) -> 14*
    if1(1,4,5) -> 14*
    if1(5,2,4) -> 14*
    if1(5,4,4) -> 14*
    if1(4,2,3) -> 14*
    if1(4,4,3) -> 14*
    f1(5) -> 11*
    f1(2) -> 11*
    f1(4) -> 11*
    f1(1) -> 11*
    f1(3) -> 11*
    false1() -> 15*
    true1() -> 15*
    c1() -> 15*
    mark1(14) -> 14,8
    mark1(11) -> 11,7
    top2(19) -> 10*
    active0(5) -> 6*
    active0(2) -> 6*
    active0(4) -> 6*
    active0(1) -> 6*
    active0(3) -> 6*
    active2(15) -> 19*
    f0(5) -> 7*
    f0(2) -> 7*
    f0(4) -> 7*
    f0(1) -> 7*
    f0(3) -> 7*
    mark0(5) -> 1*
    mark0(2) -> 1*
    mark0(4) -> 1*
    mark0(1) -> 1*
    mark0(3) -> 1*
    if0(5,4,1) -> 8*
    if0(3,4,2) -> 8*
    if0(2,1,5) -> 8*
    if0(2,3,5) -> 8*
    if0(2,5,5) -> 8*
    if0(1,1,4) -> 8*
    if0(1,3,4) -> 8*
    if0(1,5,4) -> 8*
    if0(5,1,3) -> 8*
    if0(5,3,3) -> 8*
    if0(5,5,3) -> 8*
    if0(4,1,2) -> 8*
    if0(1,1,1) -> 8*
    if0(4,3,2) -> 8*
    if0(1,3,1) -> 8*
    if0(4,5,2) -> 8*
    if0(1,5,1) -> 8*
    if0(3,2,5) -> 8*
    if0(3,4,5) -> 8*
    if0(2,2,4) -> 8*
    if0(2,4,4) -> 8*
    if0(1,2,3) -> 8*
    if0(1,4,3) -> 8*
    if0(5,2,2) -> 8*
    if0(2,2,1) -> 8*
    if0(5,4,2) -> 8*
    if0(2,4,1) -> 8*
    if0(4,1,5) -> 8*
    if0(4,3,5) -> 8*
    if0(4,5,5) -> 8*
    if0(3,1,4) -> 8*
    if0(3,3,4) -> 8*
    if0(3,5,4) -> 8*
    if0(2,1,3) -> 8*
    if0(2,3,3) -> 8*
    if0(2,5,3) -> 8*
    if0(3,1,1) -> 8*
    if0(1,1,2) -> 8*
    if0(3,3,1) -> 8*
    if0(1,3,2) -> 8*
    if0(3,5,1) -> 8*
    if0(1,5,2) -> 8*
    if0(5,2,5) -> 8*
    if0(5,4,5) -> 8*
    if0(4,2,4) -> 8*
    if0(4,4,4) -> 8*
    if0(3,2,3) -> 8*
    if0(3,4,3) -> 8*
    if0(4,2,1) -> 8*
    if0(2,2,2) -> 8*
    if0(4,4,1) -> 8*
    if0(2,4,2) -> 8*
    if0(1,1,5) -> 8*
    if0(1,3,5) -> 8*
    if0(1,5,5) -> 8*
    if0(5,1,4) -> 8*
    if0(5,3,4) -> 8*
    if0(5,5,4) -> 8*
    if0(4,1,3) -> 8*
    if0(4,3,3) -> 8*
    if0(4,5,3) -> 8*
    if0(5,1,1) -> 8*
    if0(3,1,2) -> 8*
    if0(5,3,1) -> 8*
    if0(3,3,2) -> 8*
    if0(5,5,1) -> 8*
    if0(3,5,2) -> 8*
    if0(2,2,5) -> 8*
    if0(2,4,5) -> 8*
    if0(1,2,4) -> 8*
    if0(1,4,4) -> 8*
    if0(5,2,3) -> 8*
    if0(5,4,3) -> 8*
    if0(4,2,2) -> 8*
    if0(1,2,1) -> 8*
    if0(4,4,2) -> 8*
    if0(1,4,1) -> 8*
    if0(3,1,5) -> 8*
    if0(3,3,5) -> 8*
    if0(3,5,5) -> 8*
    if0(2,1,4) -> 8*
    if0(2,3,4) -> 8*
    if0(2,5,4) -> 8*
    if0(1,1,3) -> 8*
    if0(1,3,3) -> 8*
    if0(1,5,3) -> 8*
    if0(5,1,2) -> 8*
    if0(2,1,1) -> 8*
    if0(5,3,2) -> 8*
    if0(2,3,1) -> 8*
    if0(5,5,2) -> 8*
    if0(2,5,1) -> 8*
    if0(4,2,5) -> 8*
    if0(4,4,5) -> 8*
    if0(3,2,4) -> 8*
    if0(3,4,4) -> 8*
    if0(2,2,3) -> 8*
    if0(2,4,3) -> 8*
    if0(3,2,1) -> 8*
    if0(1,2,2) -> 8*
    if0(3,4,1) -> 8*
    if0(1,4,2) -> 8*
    if0(5,1,5) -> 8*
    if0(5,3,5) -> 8*
    if0(5,5,5) -> 8*
    if0(4,1,4) -> 8*
    if0(4,3,4) -> 8*
    if0(4,5,4) -> 8*
    if0(3,1,3) -> 8*
    if0(3,3,3) -> 8*
    if0(3,5,3) -> 8*
    if0(4,1,1) -> 8*
    if0(2,1,2) -> 8*
    if0(4,3,1) -> 8*
    if0(2,3,2) -> 8*
    if0(4,5,1) -> 8*
    if0(2,5,2) -> 8*
    if0(1,2,5) -> 8*
    if0(1,4,5) -> 8*
    if0(5,2,4) -> 8*
    if0(5,4,4) -> 8*
    if0(4,2,3) -> 8*
    if0(4,4,3) -> 8*
    if0(5,2,1) -> 8*
    if0(3,2,2) -> 8*
    c0() -> 2*
    true0() -> 3*
    false0() -> 4*
    proper0(5) -> 9*
    proper0(2) -> 9*
    proper0(4) -> 9*
    proper0(1) -> 9*
    proper0(3) -> 9*
    ok0(5) -> 5*
    ok0(2) -> 5*
    ok0(4) -> 5*
    ok0(1) -> 5*
    ok0(3) -> 5*
    top0(5) -> 10*
    top0(2) -> 10*
    top0(4) -> 10*
    top0(1) -> 10*
    top0(3) -> 10*
  problem:

  Qed

Tool IRC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 C

stdout:

YES(?,O(n^1))

Tool IRC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 C

stdout:

YES(?,O(n^1))

'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer:           YES(?,O(n^1))
Input Problem:    innermost runtime-complexity with respect to
  Rules:
    {  active(f(X)) -> mark(if(X, c(), f(true())))
     , active(if(true(), X, Y)) -> mark(X)
     , active(if(false(), X, Y)) -> mark(Y)
     , active(f(X)) -> f(active(X))
     , active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
     , active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
     , f(mark(X)) -> mark(f(X))
     , if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
     , if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
     , proper(f(X)) -> f(proper(X))
     , proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
     , proper(c()) -> ok(c())
     , proper(true()) -> ok(true())
     , proper(false()) -> ok(false())
     , f(ok(X)) -> ok(f(X))
     , if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
     , top(mark(X)) -> top(proper(X))
     , top(ok(X)) -> top(active(X))}

Proof Output:
  'Bounds with minimal-enrichment and initial automaton 'match'' proved the best result:

  Details:
  --------
    'Bounds with minimal-enrichment and initial automaton 'match'' succeeded with the following output:
     'Bounds with minimal-enrichment and initial automaton 'match''
     --------------------------------------------------------------
     Answer:           YES(?,O(n^1))
     Input Problem:    innermost runtime-complexity with respect to
       Rules:
         {  active(f(X)) -> mark(if(X, c(), f(true())))
          , active(if(true(), X, Y)) -> mark(X)
          , active(if(false(), X, Y)) -> mark(Y)
          , active(f(X)) -> f(active(X))
          , active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
          , active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
          , f(mark(X)) -> mark(f(X))
          , if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
          , if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
          , proper(f(X)) -> f(proper(X))
          , proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
          , proper(c()) -> ok(c())
          , proper(true()) -> ok(true())
          , proper(false()) -> ok(false())
          , f(ok(X)) -> ok(f(X))
          , if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
          , top(mark(X)) -> top(proper(X))
          , top(ok(X)) -> top(active(X))}

     Proof Output:
       The problem is match-bounded by 2.
       The enriched problem is compatible with the following automaton:
       {  active_0(2) -> 1
        , active_1(2) -> 5
        , active_2(4) -> 6
        , f_0(2) -> 1
        , f_1(2) -> 3
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , if_0(2, 2, 2) -> 1
        , if_1(2, 2, 2) -> 3
        , c_0() -> 2
        , c_1() -> 4
        , true_0() -> 2
        , true_1() -> 4
        , false_0() -> 2
        , false_1() -> 4
        , proper_0(2) -> 1
        , proper_1(2) -> 5
        , ok_0(2) -> 2
        , ok_1(3) -> 1
        , ok_1(3) -> 3
        , ok_1(4) -> 1
        , ok_1(4) -> 5
        , top_0(2) -> 1
        , top_1(5) -> 1
        , top_2(6) -> 1}

Tool RC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 C

stdout:

YES(?,O(n^1))

Tool RC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 C

stdout:

YES(?,O(n^1))

'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer:           YES(?,O(n^1))
Input Problem:    runtime-complexity with respect to
  Rules:
    {  active(f(X)) -> mark(if(X, c(), f(true())))
     , active(if(true(), X, Y)) -> mark(X)
     , active(if(false(), X, Y)) -> mark(Y)
     , active(f(X)) -> f(active(X))
     , active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
     , active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
     , f(mark(X)) -> mark(f(X))
     , if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
     , if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
     , proper(f(X)) -> f(proper(X))
     , proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
     , proper(c()) -> ok(c())
     , proper(true()) -> ok(true())
     , proper(false()) -> ok(false())
     , f(ok(X)) -> ok(f(X))
     , if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
     , top(mark(X)) -> top(proper(X))
     , top(ok(X)) -> top(active(X))}

Proof Output:
  'Bounds with minimal-enrichment and initial automaton 'match'' proved the best result:

  Details:
  --------
    'Bounds with minimal-enrichment and initial automaton 'match'' succeeded with the following output:
     'Bounds with minimal-enrichment and initial automaton 'match''
     --------------------------------------------------------------
     Answer:           YES(?,O(n^1))
     Input Problem:    runtime-complexity with respect to
       Rules:
         {  active(f(X)) -> mark(if(X, c(), f(true())))
          , active(if(true(), X, Y)) -> mark(X)
          , active(if(false(), X, Y)) -> mark(Y)
          , active(f(X)) -> f(active(X))
          , active(if(X1, X2, X3)) -> if(active(X1), X2, X3)
          , active(if(X1, X2, X3)) -> if(X1, active(X2), X3)
          , f(mark(X)) -> mark(f(X))
          , if(mark(X1), X2, X3) -> mark(if(X1, X2, X3))
          , if(X1, mark(X2), X3) -> mark(if(X1, X2, X3))
          , proper(f(X)) -> f(proper(X))
          , proper(if(X1, X2, X3)) -> if(proper(X1), proper(X2), proper(X3))
          , proper(c()) -> ok(c())
          , proper(true()) -> ok(true())
          , proper(false()) -> ok(false())
          , f(ok(X)) -> ok(f(X))
          , if(ok(X1), ok(X2), ok(X3)) -> ok(if(X1, X2, X3))
          , top(mark(X)) -> top(proper(X))
          , top(ok(X)) -> top(active(X))}

     Proof Output:
       The problem is match-bounded by 2.
       The enriched problem is compatible with the following automaton:
       {  active_0(2) -> 1
        , active_1(2) -> 5
        , active_2(4) -> 6
        , f_0(2) -> 1
        , f_1(2) -> 3
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , if_0(2, 2, 2) -> 1
        , if_1(2, 2, 2) -> 3
        , c_0() -> 2
        , c_1() -> 4
        , true_0() -> 2
        , true_1() -> 4
        , false_0() -> 2
        , false_1() -> 4
        , proper_0(2) -> 1
        , proper_1(2) -> 5
        , ok_0(2) -> 2
        , ok_1(3) -> 1
        , ok_1(3) -> 3
        , ok_1(4) -> 1
        , ok_1(4) -> 5
        , top_0(2) -> 1
        , top_1(5) -> 1
        , top_2(6) -> 1}